# Union of Power Sets is Power Set of unions. [duplicate]

I've been working on this homework problem for a while now and I can't seem to figure it out.

The problem is as follows:

Prove that for any sets $A$ and $B$, $P(A) \cup P(B) = P(A\cup B)$ then either $A\subseteq B$ or $B\subseteq A$. I'm having trouble understanding what it means to have an equation as one of my hypotheses.

I looked online and I couldn't find anything online, so hopefully this is alright.

Thanks!

Any help would be appreciated

## marked as duplicate by Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 28 '18 at 23:42

• Equation as hypothesis: for any real number $x$, if $x^2 = 2$ then either $x = \sqrt{2}$ or $x = -\sqrt{2}$. – Patrick Stevens Feb 28 '18 at 22:45
• Probably easiest to prove by contraposition. If neither $A \subseteq B$ or $B \subseteq A$, then there are $x \in A, y \in B$ such that $x \notin B$ and $y \notin A$. Now consider the set $\{x,y\}$. – Hans Engler Feb 28 '18 at 22:48